Question

I have drawn a right angled triangle ABC whose ∠A is right angle. I took two points P and Q on the sides AB and AC respectively. By joining P,Q; B,Q and C,P let us Prove that,BQ2 + PC2 = BC2 + PQ2. ​

Answer 1

Answer:

Proof : In the figure ∆ABC,

∆ABQ, ∆APC, ∆ABC, are right angled triangle whose ∠A = 90°

∴ BC² = AB² + AC² ;

BQ² = AQ² + AB² ;

PC² = AP² + AC² ;

PQ² = AP² + AQ²

Now, BQ² + PC² = AQ²+ AB² +AP²+ AC²

and BC²+ PQ²

=> AB²+ AC²+ AP²+ AQ² = AQ² + AB²+ AP² + AC²

∴BQ² + PC² = BC² + PQ² (PROVED)

Answer 2

EXPLANATION.

In right angle triangle ΔABC.

∠A is right angled triangle.

points P and Q on the sides AB and AC.

By joining P,Q and B,Q and C,Q.

As we know that,

in right angled triangle,

⇒ H² = P² + B².

In right angled triangle ΔBQA.

⇒ (BQ)² = (AB)² + (AQ)². ⇒(1).

In right angled triangle ΔPAC.

⇒ (CP)² = (PA)² + (AC)². ⇒ (2).

Adding equation (1) & (2), we get.

⇒ BP² + CP² = AB² + AQ² + PA² + AC². ⇒(3).

In right angled triangle ΔBAC.

⇒ (BC)² = (BA)² + (AC)². ⇒(4).

In right angled triangle ΔPAQ.

⇒ (PQ)² = (PA)² + (AQ)². ⇒(5).

Solving equation (3),(4) and (5), we get.

From equation (4), we get.

⇒ BC² - BA² = AC². ⇒ (6).

From equation (5), we get.

⇒ PQ² - PA² = AQ². (7).

Put the value of equation (6) & (7) in equation (3), we get.

⇒ BQ² + CP² = AB² + (PQ² - PA²) + PA² + (BC² - BA²).

⇒ BQ² + CP² = AB² + PQ² - PA² + PA² + BC² - BA².

⇒ BQ² + CP² = BC² + PQ².

HENCE PROVED.