In $Δ A B C, ∠ B A C = 90^{\circ}$ . From A, AD is drawn perpendicular to BC. Prove that $A D^{2} = B D \times D C$ . [4 MARKS]

Open in App

Solution

Applying theorems: 2 Marks
Prrof: 2 Marks

Proof: In $Δ A D B ∠ 2 = 90^{\circ} [A D ⊥ B C] a n d ∠ 1 + ∠ 2 + ∠ 3 = 180^{\circ} \Rightarrow ∠ 1 + 90^{\circ} + ∠ 3 = 180^{\circ} \Rightarrow ∠ 1 + ∠ 3 = 90^{\circ} (i) A l s o ∠ 3 + ∠ 4 = 90^{\circ} \dots (i i) ∠ B A C = 90^{\circ}$
From (i) and (ii), we get
$∠ 1 + ∠ 3 = ∠ 3 + ∠ 4 \Rightarrow ∠ 1 = ∠ 4 a l s o, ∠ 2 = ∠ 5 [E a c h 90^{\circ}] ∴ Δ B D A \sim Δ A D C [A A s i m i l a r i t y] \frac{B D}{A D} = \frac{D A}{D C} \Rightarrow B C \times D C = A D^{2} o r A D^{2} = B D \times D C [A A s i m i l a r i t y]$

Suggest Corrections

Similar questions

In figure, $∠ B A C = 90^{\circ}$ and segment $A D ⊥ B C$ . Prove that ${A D}^{2} = B D \times D C$ . [3 MARKS]

∆ABC,AD perpendicular to BC and AD square=BD×DC prove angle BAC=90

A B C, A D

B C

A D^{2} = B D \times D C

B A C = 90^{0}

A D^{2} = B D \times D C .

∠ B A C

A B C, A D

B C

A D^{2} = B D \times D C

∠

B A C = 90^{o}

Join BYJU'S Learning Program