Sides AB and BC and Median AD of a triangle ABC are proportional to sides PQ AND PR and median PM of triangle PQR. Prove triangle AbC~trinangle PQR .
Answers
Given two triangles. ΔABC and ΔPQR in which AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
AB/PQ = BC/QR = AD/PM
To Prove: ΔABC ~ ΔPQR
Proof: AB/PQ = BC/QR = AD/PM
 AB/PQ = BC/QR = AD/PM (D is the mid-point of BC. M is the mid point of QR)
ΔABD ~ ΔPQM [SSS similarity criterion]
Therefore, ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]
∠ABC = ∠PQR
In ΔABC and ΔPQR
AB/PQ = BC/QR ———(i)
∠ABC = ∠PQR ——-(ii)
From above equation (i) and (ii), we get
ΔABC ~ ΔPQR [By SAS similarity criterion]
Hence Proved
To prove: △ABC∼△PQR
Proof:Let us extend AD to point D such that that AD=DE and PM upto point L such that PM=ML
Join B to E, C to E, & Q to L and R to L
(Image 2)
We know that medians is the bisector of opposite side
Hence BD=DC & AD=DE *By construction)
Hence in quadrilateral ABEC, diagonals AE and BC bisect each other at point D
∴ABEC is a parallelogram
∴AC=BE & AB=EC (opposite sides of a parallelogram are equal) →2
Similarly we can prove that
PQLR is a parallelogram.
PR=QL,PQ=LR (opposite sides of a parallelogram are equal) →3
Given that
PQ
AB
=
PR
AC
=
PM
AD
(frim 1)
⇒
PQ
AB
=
QL
BE
=
PM
AD
(from 2 and 3)
⇒
PQ
AB
=
QL
BE
=
2PM
2AD
⇒
PQ
AB
=
QL
BE
=
PL
AE
(As AD=DE,AE=AD+DE=AD+AD=2AD & PM=ML,PL=PM+ML=PM+PM=2PM)
∴△ABE∼△PQL (By SSS similarity criteria)
We know that corresponding angles of similar triangles are equal
∴∠BAE=∠QPL→4
Similarly we can prove that
△AEC∼△PLR
We know that corresponding angles of similar triangles are equal
∠CAE=∠RPL→5
Adding 4 and 5, we get
∠BAE+∠CAE=∠QPL+∠RPL
⇒∠CAB=∠RPQ→6
In △ABC and △PQR
PQ
AB
=
PR
AC
(from 1)
∠CAB=∠RPQ (from 6)
∴△ABC∼△PQR (By SAS similarity criteria)
Hence, proved