Sides ab and ac and median ad are properational to the sides of the

1.	Sides ab and ac and median ad are properational to the sides of the
Answer» Given : In\xa0{tex}\\Delta A B C \\text { and } \\Delta P Q R{/tex} The AD and PM are their medians,such that\xa0{tex}\\frac { A B } { P Q } = \\frac { A D } { P M } = \\frac { A C } { P R }{/tex}To prove :\xa0{tex}\\Delta A B C \\sim \\Delta P Q R{/tex}Construction : Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join CE and RN.Proof : In\xa0{tex}\\Delta A B D \\text { and } \\Delta E D C{/tex}\xa0{tex}AD=DE{/tex}\xa0{tex}\\angle A D B = \\angle E D C{/tex}\xa0(vertically opposite angles){tex}BD=DC\\text{(as AD is a median)}{/tex}{tex}\\therefore \\quad \\Delta A B D \\equiv \\Delta E D C{/tex}\xa0(By SAS congruency)or,\xa0{tex}AB=CE{/tex} (By CPCT)Similarly,\xa0PQ = RN {tex}{/tex}{tex}\\frac { A B } { P Q } = \\frac { A D } { P M } = \\frac { A C } { P R }{/tex}\xa0(Given)or,\xa0{tex}\\frac { C E } { R N } = \\frac { 2 A D } { 2 P M } = \\frac { A C } { P R }{/tex}or\xa0{tex}\\frac{CE}{RN}=\\frac{AE}{PN}=\\frac{AC}{PR}{/tex}So\xa0{tex}∆ACE \\sim ∆PRN{/tex}\xa0{tex}\\angle 3=\\angle 4{/tex}Similarly\xa0{tex}\\angle 1=\\angle 2{/tex}\xa0{tex}\\angle 1+\\angle3=\\angle2+\\angle4{/tex}So\xa0{tex}\\angle A=\\angle P\\text{ and}{/tex}\xa0{tex}\\frac{AB}{PQ}=\\frac{AC}{PR}\\text{(given)} {/tex}Hence\xa0{tex}∆ABC\\sim ∆PQR{/tex}

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