Question

Two equal circles intersected in P and Q,A straight line through P meets the circles in A and B .Prove that QA= QB
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Answer 1

Answer:

$\fbox \orange {A} \fbox \pink {N} \fbox \blue {S} \fbox \red {W} \fbox \purple {E} \fbox\gray {R}$

We know that PQ is the common chord in both the circles

So their corresponding arcs are equal

It can be written as

Arc PCQ=arc PDQ

We know that the congruent arcs have the same degree

So we get

∠QAP=∠QBP

We know that the base angles of an isosceles triangle are equal

So we get

QA=AB

Therefore, it is proved that QA=QB

hope it helps

Answer 2

Answer:

\fbox \orange {A} \fbox \pink {N} \fbox \blue {S} \fbox \red {W} \fbox \purple {E} \fbox\gray {R}

A

N

S

W

E

R

We know that PQ is the common chord in both the circles

So their corresponding arcs are equal

It can be written as

Arc PCQ=arc PDQ

We know that the congruent arcs have the same degree

So we get

∠QAP=∠QBP

We know that the base angles of an isosceles triangle are equal

So we get

QA=AB

Therefore, it is proved that QA=QB

hope it helps