Question

9. In the adjoining figure, ABCD is a parallelogram in which AB= 16 cm, BC = 10 cm and L is a point on AC such that CL: LA = 2:3. If BL produced meets CD at M and AD produced at N, prove that :

(1) ACLB - AALN

(ii) ACLM AALB.​

Answer 1

Answer:

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Answer 2

Solution :-

In ∆CLB and ∆ALN we have,

→ ∠CLB = ∠ALN {vertically opposite angles .}

→ ∠CBL = ∠ANL { since AD and BC are opposite sides of ll gm . So, AD || BC and BN is a transversal. Therefore, alternate angles are equal . }

So,

→ ∆CLB ~ ∆ALN {By AA similarity .}

Similarly, In ∆CLM and ∆ALB we have,

→ ∠CLM = ∠ALB {vertically opposite angles .}

→ ∠LCM = ∠LAB { Alternate angles .}

So,

→ ∆CLM ~ ∆ALB {By AA similarity .}

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