Chase the angles !

Geometry Level 3

In triangle $ABC$ , let $D$ be the mid-point of $BC$ . If $\angle ADB =45 ^\circ$ and $\angle ACD = 30^\circ$ , determine $\angle BAD$ (in degrees) .

The answer is 30.

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HariShankar Pv
May 3, 2014

$We\quad have\quad \angle BCE=30°,\quad we\quad get\quad \angle CBE=60°\\ D\quad is\quad the\quad midpoint\quad of\quad BC,\quad Since\quad BEC\quad is\quad a\quad right\quad triangle\\ D\quad is\quad the\quad circumcentre\quad \Rightarrow \quad DB=DE=DC.\\ Thus\quad in\quad triangle\quad BED,\angle DBE=60°\quad and\quad \angle BDE=\angle BED=60°\\ This\quad implies\quad that\quad BED\quad is\quad an\quad equilateral\quad triangle.\\ \therefore \quad EB=ED.\\ Using\quad \angle EBD=60°\quad and\quad \angle ADB=45°,\quad We\quad get\quad \angle ADE=15°\quad But\quad \\ \angle DAE=15°.\quad Thus\quad ED=EA.\\ We\quad hence\quad have\quad ED=EA=EB.\\ This\quad implies\quad that\quad E\quad is\quad the\quad circumcentre\quad of\quad triangle\quad BDA.\quad \\ \\ Thus\quad \angle BAD\quad =\quad \quad \frac { 1 }{ 2 } \angle BCD\quad =\quad \frac { 1 }{ 2 } \times 60°\quad =\quad 30°.$

BE=BD=DE=r(Equilateral triangle) also angleDAE=15 DEGREE,Hence triangle AED is equilateral, In triangle AEB,AE=BE=r,Hence anlgeBAE=45 DEGREE, Also angle DAE =15 DEGREE ,Hence Angle BAD=Angle BAE-Angle DAE=30 DEGREE

Jatindeep Singh - 7 years, 1 month ago

I think, at last line, there should be angle BED instead of angle BCD... @ @HariShankar PV

Nizam Ahmed - 7 years ago

Note: There is no need to place the text within Latex brackets, as it makes it much harder to type and read. You can see how I edited the question.

Calvin Lin Staff - 7 years, 1 month ago

Thank's ,I'll see that next time.

HariShankar PV - 7 years, 1 month ago

savvy?

swapnil rajawat - 7 years, 1 month ago

simplyg equilateral triangle!!!!!!!60-60-60

Algen Alesna - 7 years, 1 month ago

too much of thinking.................

Vighnesh Raut - 7 years, 1 month ago

did u think it urself or COPIED from rmo 2005 solution

Aneesh Kundu - 7 years, 1 month ago

triangle ACB is a 30-60-90 triangle thus angle ACB is 30 and angle ABC is x + 60 and angle BAC is y + 15, we have the 30 degree angle all we need to do is to determine the twice of 30 degrees angle and the right angle since angle EBD is already 60 degrees it is logically given that angle CBA is a 90 degree angle so 90 = x + 60 and x = 30 since we have angle ABE we can basically compute angle BAD ( 60 = y +15, y = 45 degrees) so BAD is 45 in degrees measurement

Sigmund Dela Cruz - 7 years, 1 month ago

answer is 30 degree

Anil Baloda - 7 years ago

A Former Brilliant Member
May 7, 2014

30! triangle BEC is a 30-60-90 triangle.

Let BE = 1. BC = 2 and EC = root 3. But, BD = 1 as well, since D is the mid-point.

Meaning that triangle BED must be equilateral since angle DBE is 60 and two sides on that angle = 1.

So DE = 1. From some separate angle chasing, we can find that angle EAD = 15, and angle ADE also = 15. So triangle AED is isosceles, so line AE = 1 as well.

Now we can see that triangle ABE is an isosceles right angles triangle with sides BE = 1 and AE = 1. Everything else follows from there.

Just concentrare nd solve it

Anil Baloda - 7 years ago

Chase the angles !

The answer is 30.

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