Question

In the figure ,ABCD is a parallelogram in which AB-16cm,BC-10cm 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 1.CLB ~ ALN
2.CLM~ALB

Answer 1

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 .}

Learn more :-

in triangle ABC seg DE parallel side BC. If 2 area of triangle ADE = area of quadrilateral DBCE find AB : AD show that B...

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2) In ∆ABC seg MN || side AC, seg MN divides ∆ABC into two parts of equal area. Determine the value of AM / AB

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Answer 2

Answer:

1. AAA test

2. AA test

Step-by-step explanation:

please refer to the picture attached

i hope it will help you

• ps. CSST in the picture is an abbreviation for, corresponding sides of similar triangles are in proportional

• ps. CSST in the picture is an abbreviation for, corresponding sides of similar triangles are in proportional i have also found out the dimensions of AN and CM if in case you needed it