In a triangle abc, am is the median such that it divides angle a in 1:2. If am is extended till d, angle abd= 90. Show that ad=2ac?

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Let the angle bam=x.

Then angle mac=2x as given in the data.

Let angle bma = y

So angle amc = (180-y) since angle bma and angle bmc lies on the same line.

Since ad is a median bm=mc

abd is a triangle with ab perpendicular to bd.

So;

ab= adcosx

Consider triangle bam. By law of sines we can write;

sinx/bm = siny/ab------(1)

Consider triangle amc. By law of sines we can write;

sin2x/mc = sin(180-y)/ac

We know that sin(180-y) = siny

Therefore;

sin2x/mc = siny/ac---------(2)

(1)/(2)

(sinx/sin2x)(mc/bm) = (siny/siny)(ac/ab)

It is given that bm=mc so mc/bm=1

Also sin2x = 2sinx*cosx

(sinx/(2sinx*cosx))(mc/bm) = (siny/siny)(ac/ab)

1/(2cosx) = ac/ab

ac = ab/(2cosx)

But ab = adcosx

ac = adcosx/(2cosx)

ac = ad/2

**ad = 2ac**

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