Question

In ΔABC I have drawn two perpendiculars from two vertices B and C on AC and AB (AC>AB) which are intersected each other at the point P. Let us prove that, AC2 + BP2 = AB2 + CP2​

Answer 1

Given:

$\triangle ABC$ in which perpendiculars are drawn from B and C to AC and AB.

To prove:

$AC^2 + BP^2 = AB^2 + CP^2$

Proof:

First of all, let us do the construction as per the image attached in the answer area.

The perpendiculars from B and C intersect AC and AB at D and E respectively which intersect at P.

Let us do a construction, join P with A.

Let us have a look at the Pythagorean theorem.

Following holds true in a right angled triangle:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}$

Now, let us consider LHS:

$AC^2 + BP^2$

Using Pythagorean theorem in right angled triangle $\triangle AEC, \triangle BPE$:

$AE^2 + CE^2 +PE^2 + EB^2 \\\Rightarrow AE^2 +PE^2 + EB^2 + CE^2(\triangle AEP, \triangle EBC)\\\Rightarrow AP^2 + BC^2\\\Rightarrow AD^2 +PD^2 + BD^2 + CD^2\\\Rightarrow AD^2 + BD^2 + CD^2+PD^2\ (\triangle ABD, \triangle BDC)\\\Rightarrow AB^2 + CP^2$

Which is equal to RHS (Right Hand Side).

Hence proved that:

$AC^2 + BP^2 = AB^2 + CP^2$