Question

In the given figure, point P is the centre of the A
circle. Chord AB
perpendicular chord CD. chord AB and chord CD
intersect at point M. Prove that angle PAD = angle CAB.
​

Answer 1

Answer:

AB = CD

Step-by-step explanation:

Construction : Draw perpendicular from point to the chord CD and AB

To prove : AB=CD

Proof : In △MEP and △NEP

∠EMP=∠ENP=90

∘

∠AEP=∠DEP(Given)

EP=EP(common)

Thus, by AAS congruence rule, △MEP≅△NEP

So, MP=NP(CPCT)

We know that chords equidistant from the centre are equal to each other.

Hence , AB=CD

Answer 2

Answer :-

Construction :-

Draw perpendicular from point to the chord CD and AB

chord CD and AB

To prove :- AB=CD

Proof :- In △MEP and △NEP

In △MEP and △NEP∠EMP=∠ENP=90

In △MEP and △NEP∠EMP=∠ENP=90 ∠AEP=∠DEP(Given)

EP=EP(common)

EP=EP(common)Thus, by AAS congruence rule, △MEP≅△NEP

EP=EP(common)Thus, by AAS congruence rule, △MEP≅△NEP So, MP=NP(CPCT)

EP=EP(common)

Thus, by AAS congruence rule, △MEP≅△NEP So, MP=NP(CPCT)

We know that chords equidistant from the centre are equal to each other.

EP=EP(common)

Thus, by AAS congruence rule,

△MEP≅△NEP

So, MP=NP(CPCT)

We know that chords equidistant from the centre are equal to each other.

Hence , AB=CD

Hopefully It Help You !!!