Question

Let ABCD be a parallelogram. Two points E and Fare chosen on the sides BC and Co.
respectively, such that B = m, and = n, lines AE and Bf intersect at G. Prove
that the ratio
A/G=(m+1)(n+1)/mn​

Answer 1

SOLUTION:-

Given:

ABCD is a parallelogram.

Two points E & F choose on sides BC & CO respectively.

Such that EB/EC= m & FC/FD= n.

To prove:

$\frac{AG}{GE} = \frac{(m + 1)(n - 1)}{mn}$

Proof:

Extend BF, so it intersects AD at J.

Now,

Let ∆AGJ & ∆BGE

$\angle AGJ = \angle BGE \: \: \: (vertically \: opposite \: angle) \\ \\ \angle AJG = \angle GBE \: \: \: (alternate \: interior \: angle)$

Therefore,

∆AGJ∼∆BGE [By A.A rule]

$= > \frac{AG}{GE} = \frac{AJ}{BE} \: \: \: \: [ by \: c.p.c.t]$

Now,

Consider ∆DFJ & ∆ABJ

$\angle J= \angle J\: \: \: (common) \\ \\ \angle DFJ = \angle ABJ \: \: \: \: (corresponding \: angle)$

Therefore,

∆DFJ∼∆ABJ [By A.A rule]

$= > \frac{</strong><strong>DJ</strong><strong>}{</strong><strong>AJ</strong><strong>} = \frac{</strong><strong>DF</strong><strong>}{</strong><strong>AB</strong><strong>} .........(1)$

Now,we have,

$\frac{</strong><strong>AD</strong><strong>}{</strong><strong>AJ</strong><strong>} \\ \\ = > 1 - \frac{</strong><strong>DJ</strong><strong>}{</strong><strong>AJ</strong><strong>} \\ \\ = > 1 - \frac{</strong><strong>DF</strong><strong>}{</strong><strong>AB</strong><strong>} \: \: \: </strong><strong>[</strong><strong>from</strong><strong> \: (1)</strong><strong>]</strong><strong>\\ \\ = > 1 - \frac{</strong><strong>DF</strong><strong>}{</strong><strong>CD</strong><strong>} \: \: \: (opposite \: sides \: of \: parallelogram \: are \: equal) \\ \\ = > \frac{</strong><strong>CD</strong><strong> - </strong><strong>DF</strong><strong>}{</strong><strong>CD</strong><strong>} \\ \\ = > \frac{</strong><strong>FC</strong><strong>}{</strong><strong>CD</strong><strong>} \\ \\ = > \frac{</strong><strong>AD</strong><strong>}{</strong><strong>AJ</strong><strong>} = \frac{</strong><strong>FC</strong><strong>}{</strong><strong>CD</strong><strong>} ..........(2) \\ </strong><strong>so\</strong><strong>\</strong><strong> </strong><strong>\</strong><strong>\</strong><strong> \: \: \frac{</strong><strong>AG</strong><strong>}{</strong><strong>GE</strong><strong>} = \frac{</strong><strong>AJ</strong><strong>}{</strong><strong>BE</strong><strong>} \\ \\ = > \frac{</strong><strong>AG</strong><strong>}{</strong><strong>GE</strong><strong>} = \frac{</strong><strong>AJ</strong><strong>}{</strong><strong>AD</strong><strong>} \times \frac{</strong><strong>BC</strong><strong>}{</strong><strong>BE</strong><strong>} \\ \\ = > \frac{</strong><strong>AG</strong><strong>}{</strong><strong>GE</strong><strong>} = \frac{</strong><strong>CD</strong><strong>}{</strong><strong>FC</strong><strong>} \times \frac{</strong><strong>BC</strong><strong>}{</strong><strong>BE</strong><strong>} \: \: \: </strong><strong>[</strong><strong>from</strong><strong> \: (2)</strong><strong>]</strong><strong>\\ \\ = > \frac{</strong><strong>AG</strong><strong>}{</strong><strong>GE</strong><strong>} = \frac{</strong><strong>FC</strong><strong> + </strong><strong>DF</strong><strong>}{</strong><strong>FC</strong><strong>} \times \frac{</strong><strong>BE</strong><strong> + </strong><strong>EC</strong><strong>}{</strong><strong>BE</strong><strong>} \\ \\ = > \frac{</strong><strong>AG</strong><strong>}{</strong><strong>GE</strong><strong>} = (1 + \frac{</strong><strong>FD</strong><strong>}{</strong><strong>FC</strong><strong>} ) \times (1 + \frac{</strong><strong>EC</strong><strong>}{</strong><strong>BE</strong><strong>} ) \\ \\ = > \frac{</strong><strong>AG</strong><strong>}{</strong><strong>GE</strong><strong>} = (1 + \frac{1}{n} )(1 + \frac{1}{m} ) \\ \\ = > \frac{</strong><strong>AG</strong><strong>}{</strong><strong>GE</strong><strong>} = \frac{(m + 1)(n + 1)}{mn}$

Hence, proved

Hope it helps ☺️