In given figure, considering triangles BEP and CPD, prove that

$B P \times P D = E P \times P C$

Open in App

Solution

Given a $△ A B C$ in which $B D ⊥ A C$ and $C E ⊥ A B$ and BD and CE intersect at P.

To prove $B P \times P D = E P \times P C$

Proof In $△ E P B$ and $△ D P C$ , we have

$∠ P E B = ∠ P D C$ [Each equal to $90^{\circ}$ ]

$∠ E P B = ∠ D P C$ [Vertically opposite angles]

Thus, by AA-criterion of similarity, we have

$△ E P B \sim △ D P C$

$\frac{E P}{D P} = \frac{P B}{P C}$

$\Rightarrow$ $B P \times P D = E P \times P C$ $[H e n c e p r o v e d]$

Suggest Corrections

Similar questions

B E P

C P D

B P \times P D = E P \times P C

In the given figure, AC = AE. Select the statements that are true.

In the given figure, AC = AE.

Show that :

(i) CP = EP

(ii) BP = DP.

In the figure, AB and CD are parallel.

Prove that AP × PC = BP × PD

∠ B A C = ∠ B D C = 90^{\circ}

Join BYJU'S Learning Program