Math Problem Statement

如图1,已知 CD // EF,A, B 分别是 CD 和 EF 上一点,BC 平分 ∠ABE,BD 平分 ∠ABF. (1) 证明: BD ⊥ BC; (2) 如图2,若 G 是 BF 上一点,且 ∠BAG = 50°,作 ∠DAG 的平分线交 BD 于点 P,求 ∠APD 的度数; (3) 如图3,过 A 作 AN ⊥ EF 于点 N,作 AQ // BC EF 交于点 Q,AP 平分 ∠BAN 交 BF 于 P,求 ∠PAQ 的度数。

Solution

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Let me break down the details of the problem:

  1. Part (1) asks to prove that BDBCBD \perp BC, given that CDEFCD \parallel EF, AA and BB lie on CDCD and EFEF respectively, BCBC bisects ABE\angle ABE, and BDBD bisects ABF\angle ABF.

  2. Part (2) is about finding the angle APD\angle APD, with point GG lying on BFBF and BAG=50\angle BAG = 50^\circ. The angle bisector of DAG\angle DAG intersects BDBD at point PP.

  3. Part (3) involves finding PAQ\angle PAQ, given that ANEFAN \perp EF, AQBCAQ \parallel BC, EFEF intersects BCBC at point QQ, and the angle bisector of BAN\angle BAN intersects BFBF at point PP.


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Math Problem Analysis

Mathematical Concepts

Geometric Proof
Angle Bisectors
Perpendicularity
Parallel Lines
Angle Calculations

Formulas

Angle Bisector Theorem
Perpendicularity Conditions

Theorems

Angle Bisector Theorem
Properties of Parallel Lines
Geometric Constructions

Suitable Grade Level

Grades 9-12