Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 986

## Wednesday, February 19, 2014

### Geometry Problem 986: Triangle, Median, Midpoint, Equal Angles, 45 Degrees

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ReplyDeleteWe have ∆ (BDC) similar to ∆ (ABC ) …. Case AA

So ∠(ABC)=45

Draw circumcircle of ADB, BC will tangent to this circle

∠ (AOB)=2∠ (DBC)=90

From D draw a line //OA . This line will bisect OB => ∠ (DOB)=60

So x=∠ (DAB)= ½ ∠ (DOB)=30

Non-elementary proof:

ReplyDeleteΔABC~ΔBDC

If DC=1, AC=2, then BC=√2.

By sine law,

sin x / 1 = sin 45° / √2

sin x = 1/2

x = 30°

Circumcenter O of triangle DCB has <DOB=2(180-<ACB)=2(45+x)=90+2x, so <OBD=45-x=<ABD and O is on AB . x is always less than 45 so <ACB is obtuse so O is on the same side of the line DB as segment AB.

ReplyDeleteP is on AC such that OP is perpendicular to DC. Therefore, CO^2=CP*CA=DC/2*DC*2=DC^2 because <COB=2*45=90, making triangle COD equilateral, 2x=60, x=30.

/_ABC = /_ABD + /_DBC = (45-x) + x = 45°. Draw CE perpendicular to AB. /_BCE = 45° and thus E lies on the perpendicular bisector of BC while /_CEB=2*/_CDB and thus E is the circumcentre of Tr. CDB which makes /_DEC = 2x and EC=ED. Now D is the circumcentre rt. Tr. AEC; hence ED=DC. In other words, Tr. EDC is equilateral or 2x=60° or x=30°

ReplyDeleteCH the altitude of Tr.BCD = 1/√2 ofCD. But BC=√2.CD from similar Tr.s. So BCH is a right Tr. with CH=1/2 BC. Hence x=30

ReplyDelete<C=135-x

ReplyDeleteIn triangle BDC

sinx/CD=sin(135-x)/BD

BD/CD=sin(135-x)/sinx-------(1)

<ABD=45-x

In triangle ABD

sin(45-x)/AD=sinx/BD

BD/AD=sinx/sin(45-x)----------(2)

Since AD=CD, (1)=(2)

sin(135-x)/sinx=sinx/sin(45-x)

(sinx)(sinx)=sin(135-x)sin(45-x)

(sinx)^2=[sin135cosx-cos135sinx][sin45cosx-cos45sinx]

(sinx)^2=(sin135cosx)^2-(cos45sinx)^2

(sinx)^2=[(cosx)^2-(sinx)^2]/2

3(sinx)^2=(cosx)^2

4(sinx)^2=1

sinx=1/2 or -1/2 (rej)

x=30