Bunuel wrote:
In the diagram below, triangle ABC is right-angled at B. What is the measure of ∠BAC?
(1) BD = BC and AE = AD.
(2) BE = DE.
Attachment:
The attachment 2017-07-18_1030.png is no longer available
It should be C
Statement 1: Insufficient(1) BD = BC and AE = AD.
This tells us both top and bottom triangles are isosceles but it does not provide us any information about the central triangle.
Statement 2: InsufficientThis tells us central triangle is an isosceles but it does not give out information about the angles.
Lets combined 1 and 2.
From 1, Since \(BD = BC\), we know \(∠BDC = ∠BCD\). Let us call this angle x. In the triangle \(BDC\), third angle \(∠CBD\) would be equal to \(180 - 2x\).
Since triangle ABC is right angled at B, \(∠EBD = 90 - (180 - 2x) = -90 + 2x\)
From 2, we know \(BE = DE\), hence \(∠EBD = ∠EDB = -90 + 2x\)
The third angle in the central triangle would be \(180 - 2 (-90 + 2x) = 360 - 4x\)
Lets name the two equal angles from triangle \(AED = y\)
At point D, we have
\(x + (-90 + 2x) + y = 180\)
\(3x + y = 270\) --- (1)
At point E, we have
\(360 - 4x + y = 180\)
\(4x - y = 180\) --- (2)
Solving the two equations, we get the value of y and since two angles in triangle AED are y, the third \(∠BAC\) can be calculated using \(180 - 2y\).
Edit: resizing diagram as it was too large.
Attachments
diagram.jpg [ 28.2 KiB | Viewed 1350 times ]
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