Question

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that AABC~ APQR.​

Answer 1

$\huge\star \underline{ \boxed{ \purple{Answer}}}\star$

We will be proving this by SAS congruence criteria.

Step-by-step explanation:

Given:

△ABC & △PQR

AD is the median of △ABC

PM is the median of △PQR

$\sf{\frac{AB}{PQ}\:=\:\frac{AC}{PR}\:=\:\frac{AD}{PM}\:.,.,.,.,.,.,.,.,.,.eq^n(i)}$

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

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),.,.,.,.,.,.,.,.$\sf{eq^n(ii)}$

Similarly we can prove that,

PQLR is a parallelogram.

PR=QL,PQ=LR (opposite sides of a parallelogram are equal) .,.,.,.,.,.,.,.,.,.,.,.,.,$\sf{eq^n(iii)}$

Given that

$\sf{\frac{AB}{PQ}\:=\:\frac{AC}{PR}\:=\:\frac{AD}{PM}\:.,.,.,.,.,.,.,.,.,.from\:eq^n(i)}$

$\sf{\frac{AB}{PQ}\:=\:\frac{BE}{QL}\:=\:\frac{AD}{PM}}$

$\sf{\frac{AB}{PQ}\:=\:\frac{BE}{QL}\:=\:\frac{2AD}{2PM}}$

$\sf{\frac{AB}{PQ}\:=\:\frac{BE}{QL}\:=\:\frac{AE}{PL}}$

(As AD=DE,AE=AD+DE=AD+AD=2AD & PM=ML,PL=PM+ML=PM+PM=2PM)

∴△ABE∼△PQL

We know that corresponding angles of similar triangles are equal

∴∠BAE=∠QPL ,.,.,.,.,.,.,.,.,.,$\sf{eq^n(iv)}$

Similarly we can prove that,

△AEC∼△PLR

We know that corresponding angles of similar triangles are equal

∠CAE=∠RPL.,.,.,.,.,.,.,.,.,.,.,$\sf{eq^n(v)}$

Adding eq(iv) & eq(v) we get,

∠BAE+∠CAE=∠QPL+∠RPL

⇒∠CAB=∠RPQ.,.,.,.,.,.,.,.,.,.$\sf{eq^n(vi)}$

In △ABC and △PQR

$\sf{\frac{AB}{PQ}\:=\:\frac{AC}{PR}.,.,.,.,.,.,.,.,.,.\:from\:eq^n(i)}$

∠CAB=∠RPQ (from 6)

∴△ABC∼△PQR (By SAS similarity criteria)

Hence proved.

Answer 2

Step-by-step explanation:

ides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that AABC~ APQR.

Hii

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