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Answer:
6 ||gm ABCD, AC is the diagonal AB = CD and AD = BC Consider DADC Given AB > AD ⇒ CD > AD [Since AB = CD] Recall that if two sides of a triangle are unequal, the angle opposite to the longer side is larger. Hence ∠DAC > ∠ACD → (1) But ∠ACD = ∠BAC [Alternate angles] Hence equation (1) becomes, ∠DAC > ∠BAC That is ∠BAC < ∠DAC
8 Since ABCD is a parallelogram. Therefore, DC∥AB.
Now DC∥AB and transversal BD intersects them at B and D.
Therefore, ∠ABD=∠BDC ∣ Alternate interior angles
in △APB and △CQD, we have
∠ABP=∠QDC ....(alternate interior angles of parallelogram ABCD and DC∥AB)
∠APB=∠CQD ....[each angle 90
∘
]
and AB=CD ∣ Opposite. sides of a || gm
Therefore, by AAS criterion of congruence
△APB≅△CQD
(ii) Since △APB≅△CQD
Therefore, AP=CQ ∣ Since corresponding parts of congruent triangles are equal
Step-by-step explanation:
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