Question

Two right triangles ABC & DBC are drawn on the same hypotenuse BC and on the same side of BC. if AC & BD intersect at P , prove that : AP * PC=BP * PD

Answer 1

Answer:

AP * PC = BP * DP

Step-by-step explanation:

ΔABC, ΔDBC are right angle triangles at A and D respectively.ΔDPC is a right angle Δ at D.

We know that,

There is a theorem to solve this type of question called Pythagoras theorem.

We have to apply it here,

$\sf =>CD^2 = CP^2 -DP^2\\\ => (CA + AP)^2 - (DB + BP)^2\\\ => CA^2 + AP^2+ 2 CA AP - DB^2 - BP^2 - 2 DB BP\\\ => CD^2+ DB^2= CA^2 - (BP^2 -AP^2) + 2 CA * AP - 2 DB BP\\\ => CB^2 = CA^2 - AB^2 + 2 CA AP - 2 DB BP\\\ => 2 AB^2 = 2 CA * AP - 2 DB * BP\quad\;as\;CB^2 = CA^2 + AB^2\\\ => AB^2 = (PC - AP) AP - (DP - BP) BP\\\ =>AP* PC - AP^2 - DP * BP + BP^2\\\ => AP,PC - DP * BP + AB^2\\\ => AP*PC = DP * BP\\\\\ Hence,Proved!$

Note:

Refer The Attached file for the figure.

Answer 2

Answer:

In ∆APB & ∆DPC,

∠A =  ∠D = 90°

∠APB = ∠DPC

[ vertically opposite angles]

∆APB  ~ ∆DPC

[By AA Similarity Criterion ]

AP/DP = PB/ PC

[corresponding sides of similar triangles are proportional]

AP × PC = BP × PD

Hence, proved.