+816 Triangles ABC and ABD are isosceles with \(AB=AC=BD\), and \(\overline{BD}\) intersects \(\overline{AC}\) at \(E\). If \(\overline{BD}\perp\overline{AC}\), then what is the value of \(\angle C+\angle D\)?
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+128579 Angle C = ACB Angle D = ADB
BAD = ADB ABC = ACB
DBA + BAC = 90 ⇒ DBA = 90 - BAC
ACB + DBC = 90 ⇒ DBC = 90 - ACB
DAC + ADB = 90 ⇒ DAC = 90 - ADB
DAC + BAC = BAD DBA + DBC = ABC
DAC + BAC = ADB DBA + DBC = ACB
[90 - ADB] + BAC = ADB DBA + [90 - ACB] = ACB
90 + BAC = 2ADB (1) 90 + DBA = 2ACB
90 + [90 - BAC] = 2ACB
180 - BAC = 2ACB
BAC = 180 - 2ACB (2)
Sub (2) into (1)
90 + [180 - 2ACB] = 2ADB
270 = 2ACB + 2ADB
135 = ACB + ADB
135 = ∠C + ∠ D
+26367 Triangles ABC and ABD are isosceles with
\(AB=AC=BD\),
and\( \overline{BD}\) intersects\( \overline{AC}\) at \(E\).
If \(\overline{BD}\perp\overline{AC}\), then what is the value of \(\angle C+\angle D\)?
\(\angle D = BDA = DAB \\ \angle C = BCA = CBA \\ \angle \alpha = DBA\\ \angle \beta = BAC \\ \angle \alpha +\angle \beta = 90^{\circ}\)
\(\begin{array}{|rcll|} \hline \angle \alpha &=& 180^{\circ} - 2\times \angle D \\ \angle \beta&=& 180^{\circ} - 2\times \angle C \\ \angle \alpha +\angle \beta &=& 180^{\circ} - 2\times \angle D+180^{\circ} - 2\times \angle C \quad & | \quad \angle \alpha +\angle \beta = 90^{\circ}\\ 90^{\circ} &=& 180^{\circ} - 2\times \angle D+180^{\circ} - 2\times \angle C \\ 90^{\circ} &=& 360^{\circ} - 2\times (\angle C + \angle D) \\ 2\times(\angle C + \angle D) &=& 360^{\circ} - 90^{\circ} \\ 2\times(\angle C + \angle D) &=& 270^{\circ} \\ \mathbf{ \angle C + \angle D } & \mathbf{=}& \mathbf{135^{\circ}} \\ \hline \end{array}\)
