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.
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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.
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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
[tex]\sf{\frac{AB}{PQ}\:=\:\frac{AC}{PR}\:=\:\frac{AD}{PM}\:.,.,.,.,.,.,.,.,.,.eq^n(i)}[/tex]
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),.,.,.,.,.,.,.,.[tex]\sf{eq^n(ii)}[/tex]
Similarly we can prove that,
PQLR is a parallelogram.
PR=QL,PQ=LR (opposite sides of a parallelogram are equal) .,.,.,.,.,.,.,.,.,.,.,.,.,[tex]\sf{eq^n(iii)}[/tex]
Given that
[tex]\sf{\frac{AB}{PQ}\:=\:\frac{AC}{PR}\:=\:\frac{AD}{PM}\:.,.,.,.,.,.,.,.,.,.from\:eq^n(i)}[/tex]
[tex]\sf{\frac{AB}{PQ}\:=\:\frac{BE}{QL}\:=\:\frac{AD}{PM}}[/tex]
[tex]\sf{\frac{AB}{PQ}\:=\:\frac{BE}{QL}\:=\:\frac{2AD}{2PM}}[/tex]
[tex]\sf{\frac{AB}{PQ}\:=\:\frac{BE}{QL}\:=\:\frac{AE}{PL}}[/tex]
(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 ,.,.,.,.,.,.,.,.,.,[tex]\sf{eq^n(iv)}[/tex]
Similarly we can prove that,
△AEC∼△PLR
We know that corresponding angles of similar triangles are equal
∠CAE=∠RPL.,.,.,.,.,.,.,.,.,.,.,[tex]\sf{eq^n(v)}[/tex]
Adding eq(iv) & eq(v) we get,
∠BAE+∠CAE=∠QPL+∠RPL
⇒∠CAB=∠RPQ.,.,.,.,.,.,.,.,.,.[tex]\sf{eq^n(vi)}[/tex]
In △ABC and △PQR
[tex]\sf{\frac{AB}{PQ}\:=\:\frac{AC}{PR}.,.,.,.,.,.,.,.,.,.\:from\:eq^n(i)}[/tex]
∠CAB=∠RPQ (from 6)
∴△ABC∼△PQR (By SAS similarity criteria)
Hence proved.
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.
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