Given quadrilateral ABCD ~ quadrilateral JKLM and AD = 14, JM = 6, and KL = 9, what is BC?
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<u>Given:</u> ∠XAM = ∠YBM = 90° and AM = BM
<u>To prove:</u> BX ≅ AY
<u>Proof:</u>
In the triangle XAM and BYM:
- ∠3 = ∠4 <em>[vertically opposite angles]</em>
- AM = BM <em> [Given]</em>
- ∠XAM = ∠YBM <em>[Both are 90°]</em>
From the ASA rule of congruency, we can say that: ΔXAM ≅ ΔYBM
Since ΔXAM ≅ ΔYBM: We can say that XM ≅ MY <em>[CPCT - Common Part of Congruent triangles]</em>
In the Triangle AMY and XMB:
- ∠1 = ∠2 <em> [vertically opposite angles]</em>
- AM = BM <em>[Given]</em>
- XM = MY <em>[Proved above]</em>
From the SAS rule of congruency, we can say that: ΔAMY ≅ ΔXMB
Since ΔAMY ≅ ΔXMB: We can finally say that BX ≅ AY <em>[CPCT]</em>
Hence Proved!
