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Question Number 6635 by Rasheed Soomro last updated on 07/Jul/16

Commented by Rasheed Soomro last updated on 09/Jul/16

In other words: The median through A in △ABC is equidistant from B and C. Or The median through a vertex of of a triangle is equidistant from its two other vertices.

Inotherwords:ThemedianthroughAin△ABCisequidistantfromBandC.OrThemedianthroughavertexofofatriangleisequidistantfromitstwoothervertices.

Commented by Rasheed Soomro last updated on 07/Jul/16

In triangle ABC , D is the midpoint of BC AD^(↔) is a median-line. BE⊥AD^(↔) and CF⊥AD^(↔) Prove that ∣BE∣=∣CF∣.

IntriangleABC,DisthemidpointofBCAD↔isamedian−line.BE⊥AD↔andCF⊥AD↔Provethat∣BE∣=∣CF∣.

Answered by Yozzii last updated on 07/Jul/16

∠ADC=∠BDE since ∠ADC and ∠BDE are vertically opposite angles by virtue of the intersection of lines AE and BC. Also, ∣DC∣=∣BD∣>0 since line AD is a median of △ABC. Since ∠DFC=∠BED=90°, △DFC and △BED are right−angled triangles. Let θ=∠ADC=∠FDC (F lies between A and D on AD). Therefore, in △DFC, sinθ=((∣FC∣)/(∣DC∣)) and, in △BED, sinθ=((∣BE∣)/(∣BD∣)). Equating both expressions for sinθ we have ((∣FC∣)/(∣DC∣))=((∣BE∣)/(∣BD∣)). But, ∣DC∣=∣BD∣. ∴ ∣FC∣=∣BF∣ ■

∠ADC=∠BDEsince∠ADCand∠BDEareverticallyoppositeanglesbyvirtueoftheintersectionoflinesAEandBC.Also,∣DC∣=∣BD∣>0sincelineADisamedianof△ABC.Since∠DFC=∠BED=90°,△DFCand△BEDareright−angledtriangles.Letθ=∠ADC=∠FDC(FliesbetweenAandDonAD).Therefore,in△DFC,sinθ=∣FC∣∣DC∣and,in△BED,sinθ=∣BE∣∣BD∣.Equatingbothexpressionsforsinθwehave∣FC∣∣DC∣=∣BE∣∣BD∣.But,∣DC∣=∣BD∣.∴∣FC∣=∣BF∣◼

Commented by Rasheed Soomro last updated on 07/Jul/16

The triangles can also be proved congruent using A.S.A≊ A.S.A theorm.

ThetrianglescanalsobeprovedcongruentusingA.S.A≊A.S.Atheorm.

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