Qno14:- Sides AB and AC and median AD of a ∆ABC are respectively proportional to sides PQ and PR and median PM of another ∆PQR۔ Show that ∆ABC ~PQR۔

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Qno14:- Sides AB and AC and median AD of a ∆ABC are respectively  proportional to sides PQ and PR and median PM of another ∆PQR۔ Show that ∆ABC ~PQR۔
Given: AB/PQ=AC/PR=AD/PM۔
and In ∆ABC, AD is the median.
• BD = DC
Also in ∆PQR, PM is the median.
• QM= MR۔

Construction: in ∆ABC Draw DG || AC and DE || AB۔
Also in ∆PQR Draw FM || PR and HM|| PQ۔
To Prove:- ∆ABC ~ ∆PQR۔
Proof:-  Since D is the mid point of BC and DG || AC.
So by Converse of Mid Point Theorem.
The converse of midpoint theorem states that: "If a line segment is drawn passing through the midpoint of any one side of a triangle and parallel to another side, then this line segment bisects the remaining third side.
 i,e DG bisects AB.
=> AG = BG
Similarly  AE = EC
Also in ∆PQR ; PF = FQ  and PH = HR
Now in ∆ABC ; GD || AE and AG|| DE
=> Quadrilateral AGDE is a Parallelogram .
Then  GE = AE ..........1.
( Opposite sides of ||gm are equal)

similarly in ∆PQR ; FM||PH & PF || HM
So Quadrilateral FMPH is a Parallelogram .
Then FM=PH............2.
Now in ∆AGD and ∆PFM
AB/PQ =AD/PM=AC/PR
i,e 2AG/2PE=AD/PM= 2AE/2PH
AG/PE=AD/PM=AE/PH .  (Cancelling 2)
Again AG/PE=AD/PM=GE/FM       (using 1 and 2).

Hence ∆AGD ~ ∆PFM (By SSS Similarity)
=>  <1   = <2 ..............3
Also <3 = <4..…..........4
Adding  3 and 4 we have.

 <1  + <3 = <2 + < 4.
<A = <P
Now in ∆ ABC and ∆PQR
AB/PQ = AC/PR . . (Given)
& <A = <P . .  . .  .  (Proved)
Hence ∆ABC ~ ∆PQR ( By SAS Similarity Criterion )
Hence Proved.


Answer is at my blog..
mathlab98.blogspot.com