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1.

ADandBEare respectively altitudes oftriangle ABCsuchthatAE=BD.ProvethatAD=BE.

Answer» AD=BE
`/_AEB and /_BDA`
AB=BA(hypotenuse)
AE=BD(given)
`/_AEB cong /_BDA(HL)`
`EB=DA`
`BE=AD`.
Discussion
2.

Prove that the angle between internal bisector ofone base angle and the external bisector of the other base angle of atriangle is equal to one half of the vertical angle.

Answer» To prove- `/_E=1/2/_A`
By exterior angle theorem
`/_A+/_B=/_ACD`
`1/2/_A+1/2/_B=1/2/_ACD`
`1/2/_A+/_1=/_2`
`/_2=/_1+1/2/_A-(1)`
`In /_BCE`
`/_ECD=/_1+/_E`
`/_2=/_1+/_E-(2)`
From equation 1 and 2
`/_1+/_E=/_1+1/2/_A`
`/_E=1/2/_A`.
Discussion
3.

If perpendiculars from any point with an angle onits arms are congruent, prove that it lies on the bisector of that angle.

Answer» To prove-`/_1=/_2`
Proof-`In /_PMB and /_PMB`
PM=PN(given)
BP=BD(common)
`/_PMB=/_PNB(each 90^@)`
`/_PMB cong /_PNB(RHS)`
`/_1=/_2`
BP bisects `/_B`.
Discussion
4.

AD, BEandCF, the altitudes oftriangle ABCare equal. Prove thatABCis an equilateral triangle.

Answer» `In /_ADC and /_BEC`
`/_D=/_E`(each 90)
AD=BE(given)
`/_C=/_C`(common)
`/_ADC cong /_BEC)(A A S)`
`AC=BC`-(1)
`In /_ADB and /_CFB`
`/_D=/_F`(each 90)
`AD=CF`(given)
`/_B=/_B`(common)
`/_ADB cong /_CFB`(AAS)
AB=CB-(2)
from equation 1 and 2
`AB=BC=AC`
Hence,`/_ABC` is equilateral.
Discussion
5.

If the bisector of the exterior vertical angle ofa triangle be parallel to the base. Show that the triangle is isosceles.

Answer» from the diagram we can see that
`/_B=/_C=x`
`AC=AB`
The triangle is hence isosceles.
Discussion
6.

In Figure, `A C=B C ,/_D C A=/_E C B`and `/_D B C=/_E A Cdot`Prove that triangles `D B C`and `E A C`are congruent, and hece `D C=E C`and `B D=A E`

Answer» Given-`AC=BC`,`/_1=/_2`,`/_3=/_4`
Proof-`/_1=/_2`
`/_1+/_DCE=/_2+/_DCE`
`/_ACE=/_DCB`
`In /_DBC and /_EAC`
`AC=BC`
`/_ACE=/_BCD`
`/_3=/_4`
`/_DBC cong /_EAC`(ASA)
`DC=ECand BD=EA.
Discussion
7.

The sides `B C ,C A`and `A B`of a triangle `A B C ,`are produced in order, forming exterior angles `/_A C D ,/_B A E`and `/_C B F`. Show that `/_A C D+/_B A E+/_C B F=360^0`

Answer» `x+y+z=180^@`
`x+/_BAE=180^@`
`y+/_CBF=180^@`
`z+/_ACD=180^@`
`(x+y+z)+/_BAE+/_CBF+/_ACD=540^@`
`/_BAE+/_CBF+/_ACD=540-180=360^@`.
Discussion
8.

`If`D is the mid-point of the hypotenuse `A C`of a right triangle `A B C`, prove that `B D=1/2A C`.GIVEN : A `A B C`in which `/_B=90^0`and `D`is the mid-point of `A Cdot`TO PROVE : `B D=1/2A C`CONSTRUCTION Produce `B D`to `E`so that `B D=D Edot`Join `E Cdot`

Answer» `In /_ADB and /_CDE`
`AD=CD`(O is midpoint of AC)
`BD=DE`(given)
`/_ADB=/_CDE`(UDA)
`/_ADB cong /_CDE(SAS)`
`EC=AB and /_CED=/_ABP`
`In /_ABC and /_ECB`
`AB=EC`(proved above)
`BC=BC`(common)
`/_ABF=/_ECB`(90)
`/_ABC cong /_ECB(SAS)`
`AC=BE`
`1/2AC=1/2BE`
`1/2AC=BD`.
Discussion
9.

In an isosceles triangle altitude from the vertexbisects the base.GIVEN : An isosceles triangle `A B C`such that `A B=A C`and an altitude `A D`from `A`on side `B Cdot`TO PROVE : `D`bisects `B C`i.e. `B D=D Cdot`Figure

Answer» Given- `AB=AC`
TO prove= BD=CD
Proof-`In /_ADB and /_ACD`
`/_ADB=/_ADC`(each 90)
`AD=AD`(common)
`/_B=/_C`
`/_ADB cong /_ACD`(AAS)
BD=CD.
Discussion
10.

If the altitude from one vertex of a trianglebisects the opposite side, then the triangle is isosceles.GIVEN : A `A B C`such that the altitude `A D`from `A`on the opposite side `B C`bisects `B C`i.e., `B D=D Cdot`TO PROVE : `A B=A C`i.e. the triangle `A B C`is isosceles.

Answer» `BO=DC`
`In /_ABD and /_ACD`
`AD=AD`(common)
`/_ADB=/_ADC`(each 90)
`BD=CD`(given)
`/_ABD cong /_ACD`(SAS)
`AB=AC`
ABC is isosceles.
Discussion
11.

If `A B C`is an isosceles triangle such that `A B=A C`and `A D`is an altitude from `A`on `B C`. Prove that(i) `/_B=/_C`(ii) `A D`bisects `B C`(iii) `A D`bisects `/_A`

Answer» ` In /_ADC and /_ABD`
`AB=AC`(given)
`/_ADB=/_ADC(each 90^@)`
`AD=AD`(common)
`/_ADc cong /_ABD(RHS)`
`/_B=/_C,BD=CD`
`/_BAD=/_CAD`.
Discussion
12.

In Figure, diagonal `A C`of a quadrilateral `A B C D`bisects the angles `A`and `C`. Prove that `A B=A D`and `C B=CD`

Answer» Given- `/_BAC=/_DAC` and `/_BCA=/_DCA`
To prove- `AB=AD and CB=CD`
Proof- `In /_ABC and /_ADC`
`/_BAC=/_DAC`(given)
`/_BCA=/_DCA`(given)
`AC=AC`(common)
`/_ABC cong /_ADC(ASA)`
`AB=AD and CB=CD`.
Discussion
13.

`P`is a point equidistant from two lines `l`and `m`intersecting at a point `A`see in figure, Show that `A P`bisects the angle between them.Figure

Answer» `In /_PAB and /_PCB`
PB=PC
AP=AD
`/_PBA=/_PCA`
`/_PAB cong /_PCB(RHS)`
`/_PAB=/_CAP`.
Discussion
14.

If two isosceles triangles have a common base,prove that the line joining their vertices bisects them at right angles.

Answer» Too prove=AD bisects BBC at right angles
Proof=2N `/_ABC ans /_ADC`
AB=AC(given)
BD=CD(given)
BC=BC(common)
`/_ABC cong /_ADC(S S S)`
`/_1=/_2`
`In /_ABE and /_ACE`
AB=AC
`/_a=/_2`
AE=AE(common)
`/_ABE cong /_ACE(SAS)`
BE=CE
`/_3=/_4`
`/_3+/_4=180^@`
`2/_3=180^@`
`/_3=90^@`.
Discussion