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In a δAbc, Bd is the Median to the Side Ac, Bd is Produced to E Such that Bd = De. Prove That: Ae is Parallel to Bc. - Mathematics

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प्रश्न

In a ΔABC, BD is the median to the side AC, BD is produced to E such that BD = DE.
Prove that: AE is parallel to BC.

योग

उत्तर

Given: A(ΔABC) in which BD is the median to AC.
BD is produced to E such that BD = DE,
We need to prove that AE II BC.
Construction: Join AE

Proof:
AD = DC          ...[ BD is median to AC ] ...(1)
In ΔBDC and ΔADE,
BD = DE                      ...[ Given ]
∠BDC = ∠ADE = 90°   ...[ Vertically opposite angles ]
AD = DC                     ...[ from(1) ]
∴ By Side-Angle-Side Criterion of congruence,
ΔBDC ≅ ΔADE
The corresponding parts of the congruent triangles are congruent.
∴ ∠EAD = ∠BCD     ...[ c.p.c.t. ]
But these are alternate angles and AC is the transversal.
Thus, AE || BC.

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Criteria for Congruence of Triangles
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Triangles [Congruency in Triangles] - Exercise 9 (B) [पृष्ठ १२५]

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सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 9 Triangles [Congruency in Triangles]
Exercise 9 (B) | Q 4 | पृष्ठ १२५
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