Question

14.In the figure, BL is perpendicular to AC and MC perpendicular to LN and AL = CN and BL = CM prove that ∆ ABC ≅ ∆ NML.

Answer 1

Answer:

in the figure the two triangles are ABC and NML

If you see in the figure carefully you'll see that LBCM is a parellelogram so, according to the properties of parellelogram ML should be equal to BC.

The next step includes ML and BC,according to the properties of parellelogram ML should be parellel to BC so LC is the transversal, hence, angle MLC and angle LCB are equal.

Next, you'll see that LC is common and ATQ, AL=CN so, if AL+LC = CN+LC hence, AC = LN. The next step includes ML and BC,according to the properties of parellelogram.

So, we have,

ML=BC

AC=LN

angle MLC=angle LCB

HENCE PROVED THAT ∆ABC ≅ ∆NML

Answer 2

Answer:

Step-by-step explanation:

In the figure the two triangles are ABC and NML

If you see in the figure carefully you'll see that LBCM is a parellelogram so, according to the properties of parellelogram ML should be equal to BC.

The next step includes ML and BC,according to the properties of parellelogram ML should be parellel to BC so LC is the transversal, hence, angle MLC and angle LCB are equal.

Next, you'll see that LC is common and ATQ, AL=CN so, if AL+LC = CN+LC hence, AC = LN. The next step includes ML and BC,according to the properties of parellelogram.

So, we have,

ML=BC

AC=LN

angle MLC=angle LCB

HENCE PROVED THAT ∆ABC ≅ ∆NML

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