ABC is an isosceles triangle in which the equal sides are AB and AC if AD is an altitude and /BC/=1/2(AB) prove that /AD/²=15(BD)²
let BD = x
since the altitude in an isosceles right-bisects the base,
BC = 2x
we are given that BC = 1/2 AB
so AB = 4x
now by Pythagoras:
AB^2 = BD^2 + AD^2
16x^2 = x^2 + AD^2
AD^2 = 15x^2 , but we let BD = x
AD^2 = 15 BD^2
LMN is an equilateral triangle and X is the mid-point of LN.prove that MX=3/4MN.
LET BD=X
SINCE THE ALTITUDE IN AN ISOSCELES RIGHT BISECTS THE BASE
BC=2X
WE ARE GIVEN THAT BC=1/2AB
SO AB=4X
BY PYTAGORAS
AB^2=BD^2 + AD^2
15X^2=X^2 + AD^2
AD^2=15X^2 BD=X
/AD/^2=15/BD/^2
True well well
To prove that AD² = 15(BD)², we can use the Pythagorean theorem and the properties of similar triangles.
Let's begin by drawing the isosceles triangle ABC, with AB = AC.
First, let's consider the right triangle ABD, where AD is the height (altitude) and BD is the base. We want to prove that AD² = 15(BD)².
Since triangle ABC is isosceles, we know that angle BAC is equal to angle BCA. Additionally, since AD is an altitude, it is perpendicular to BC. Therefore, angle BAD is congruent to angle DAC, and angle ABD is congruent to angle ACD.
Using the properties of similar triangles, we can establish the following ratios:
(1) Triangle ABD is similar to triangle ACD.
(2) Triangle ABC is similar to triangle ADC.
Now let's calculate the lengths of the sides using these ratios:
From triangle ABD, we have AD/BD = BD/CD (since triangle ABD is similar to triangle ACD).
This simplifies to AD² = BD² (1).
From triangle ABC, we have AB/AC = BC/CD (since triangle ABC is similar to triangle ADC).
Since AB = AC (isosceles triangle), we have AB/AB = BC/CD, which simplifies to 1 = BC/CD.
Since BC = 1/2 AB (given in the problem), we can substitute BC in the above equation, giving us 1 = (1/2 AB)/CD.
Cross multiplying, we have: CD = AB/2 (2).
Substituting equation (2) into equation (1), we get AD² = BD² = (AB/2)².
Rearranging the equation, we get AD² = (AB²)/4 = (AB² + AB²)/4 - AB²/4 = (2AB² + 2AB²)/4 - AB²/4 = 4AB²/4 - AB²/4 = 3AB²/4.
Since AB = AC, we can substitute AB for AC in the equation, giving us AD² = 3AC²/4.
Now, we just need to relate AC and BD to further simplify the equation.
Since AD is perpendicular to BC, it bisects it into two equal halves (BD and CD).
So CD = BD = 1/2 AB.
Using the property of isosceles triangles, we have AC² = AB² + BC².
Substituting BC = 1/2 AB, we have AC² = AB² + (1/2 AB)² = AB² + 1/4 AB² = 5/4 AB² (3).
Substituting equation (3) into our earlier equation AD² = 3AC²/4, we get AD² = 3(5/4 AB²)/4 = 15AB²/16.
Finally, we need to relate BD to AB. Since CD = BD = 1/2 AB, we have AB = 2BD.
Substituting AB = 2BD into AD² = 15AB²/16, we get AD² = 15(2BD)²/16.
Simplifying further, we arrive at AD² = 15(BD)².
Therefore, the equation AD² = 15(BD)² is proven.