Prove that the altitude to the base of an isosceles triangle is also the angle bisector of the angle from which it is drawn
1 answer:
Step-by-step explanation:
To prove- Altitude to the base of an isosceles triangle bisects the angle from which it is drawn.
In the given figure ΔABC (isosceles triangle) whose sides AB=AC
From ∠A an altitude is drawn to base (BC) which meets BC at X
∠AXB=90 and ∠AXC =90
∴We need to prove ΔXAB= ΔXAC for altitude to bisect the ∠ BAC
In Δ AXB and Δ AXC
AB=AC (same hypotenuse)
AX= common altitude (serve as the common leg of the hypotenuse)
ΔAXB ≅ ΔAXC
Hence, we can conclude that the corresponding angle∠ XAB = ∠XAC. QED
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Answer:
Step-by-step explanation:
Given
Midpoint, T
Required
Determine PT
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Since T is the midpoint, then:
This implies
Solve for x
Length PQ can then be calculated using:
Collect Like Terms
Substitute 6 for x
Answer:
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units and shade inside the V
Step-by-step explanation:
we know that
The function has the vertex at point
The function has the vertex at point
so
the rule of the translation is
That means
The translation is units up
The solution of the inequality
is the shaded area inside the V
see the attached figure to better understand the problem
therefore
the answer is
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units and shade inside the V
