In a right triangle ABC, right angled at B, D is a point on hypotenuse

1.	In a right triangle ABC, right angled at B, D is a point on hypotenuse such that BD⊥AC, if DP⊥AB and DQ⊥BC then prove that (a) DQ2 = Dp.QC. (b) DP2 = DQ .AP2
Answer» We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other. (a) Now using the same property in In ∆BDC, we get ∆CQD ~ ∆DQB CQ/DQ = DQ/QB ⇒ DQ² = QB. CQ Now. Since all the angles in quadrilateral BQDP are right angles. Hence, BQDP is a rectangle. So, QB = DP and DQ = PB ∴ DQ²= DP.CQ (b) Similarly, ∆APD ~ ∆DPB AP/DP = PD/PB ⇒ DP² = AP.PB ⇒ DP²= AP.DQ [∵ DQ = PB]

In a right triangle ABC, right angled at B, D is a point on hypotenuse such that BD⊥AC, if DP⊥AB and DQ⊥BC then prove that (a) DQ2 = Dp.QC. (b) DP2 = DQ .AP2

Answer»

We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other.

(a) Now using the same property in In ∆BDC, we get

∆CQD ~ ∆DQB

CQ/DQ = DQ/QB

⇒ DQ² = QB. CQ

Now. Since all the angles in quadrilateral BQDP are right angles.

Hence, BQDP is a rectangle.

So, QB = DP and DQ = PB

∴ DQ²= DP.CQ

(b) Similarly, ∆APD ~ ∆DPB

AP/DP = PD/PB

⇒ DP² = AP.PB

⇒ DP²= AP.DQ [∵ DQ = PB]

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