24. The interior angles of a polygulis 80. Find the number of sides of

1.	24. The interior angles of a polygulis 80. Find the number of sides of the polygon25.In triangle ABC, D is the mid-point of BC and AE t-BC. If AC > AB, then show that:AB2 = AD2BC2
Answer» PROOF: [for easy convenience, let me denote all squares by * and multiplication by 'x'] 1) Since AE is perpendicular to BC, ABE is a right triangle with angle E = 90 deg. 2) Hence applying Pythagoras theorem we have, AB* = AE* + BE* 3) In similar way from the right triangle, ADE, AD* = AE* + DE, => AE = AD* - DE* 4) Substituting for AE* from step 3 in step 2, we have, AB* = AD* - DE* + BE* 5) From the figure (Kindly make according to the description given in data), BE = BD - DE 6) Substituting this in (4), AB* = AD* - DE* + (BD-DE)* 7) Expanding, AB* = AD* - DE* + BD* + DE* - 2 x BD x DE 8) Cancelling -DE* and +DE* and substituting BD = (1/2)BC {Since, D is mid point of BC} we get, AB* = AD* + (1/4)BC* - BC x DE Thus it is proved that "AB^2=AD^2-BC*DE+(1/4)BC^2"

24. The interior angles of a polygulis 80. Find the number of sides of the polygon25.In triangle ABC, D is the mid-point of BC and AE t-BC. If AC > AB, then show that:AB2 = AD2BC2

Answer»

PROOF: [for easy convenience, let me denote all squares by * and multiplication by 'x']

1) Since AE is perpendicular to BC, ABE is a right triangle with angle E = 90 deg.

2) Hence applying Pythagoras theorem we have, AB* = AE* + BE*

3) In similar way from the right triangle, ADE, AD* = AE* + DE*, => AE* = AD* - DE*

4) Substituting for AE* from step 3 in step 2, we have,

AB* = AD* - DE* + BE*

5) From the figure (Kindly make according to the description given in data),

BE = BD - DE

6) Substituting this in (4), AB* = AD* - DE* + (BD-DE)*

7) Expanding, AB* = AD* - DE* + BD* + DE* - 2 x BD x DE

8) Cancelling -DE* and +DE* and substituting BD = (1/2)BC

{Since, D is mid point of BC}

we get, AB* = AD* + (1/4)BC* - BC x DE

Thus it is proved that "AB^2=AD^2-BC*DE+(1/4)BC^2"