Question

In a triangle ABC . P and q are the mid points of CA and CB. If angle C is 90°. Then show that 4AQ² = 4AC² + BC²

Answer 1

In triangle ABC the right angle is C, the sides are AC and BC, the hypotenuse is AB. Q is the midpoint of BC, so (CQ) = 1/2 (BC). And so (CQ)^2 = 1/4 (BC)^2 or (BC)^2 = 4(CQ)^2

By Pythagoras in triangle ACQ, (AQ)^2 = (CQ)^2 + (AC)^2

(CQ)^2 = (AQ)^2 - (AC)^2

4(CQ)^2 = 4(AQ)^2 - 4(AC)^2

By Pythagoras in triangle ABC: (AB)^2 = (BC)^2 + (AC)^2 then substituting:

(AB)^2 = 4(CQ)^2 + (AC)^2

(AB)^2 = 4(AQ)^2 - 4(AC)^2 + (AC)^2 = 4(AQ)^2 - 3(AC)^2

I hope that help you

Answer 2

hey mate here is your answer....

By Pythagoras theorem.

AC²+QC²=AC²

NOW,

MULTIPLY BY 4

4AC² + 4QC²= 4AC²

'Q' IS the mid point of 'BC' so bc = 2qc

4ac²+4(bc/2)²=4ac²

4ac²+ bc² =4ac²

hope it will help you

plz mark as brainliest